A355713 Numbers k such that k and k+1 have the same sum of 5-smooth divisors.

175, 2224, 2575, 4975, 7024, 9424, 9775, 11824, 12175, 14224, 14575, 16975, 19024, 21424, 21775, 23824, 24175, 26224, 26575, 28975, 31024, 33424, 33775, 35824, 36175, 38224, 38575, 40975, 43024, 45424, 45775, 47824, 48175, 50224, 50575, 52975, 55024, 57424, 57775

Offset: 1

Author

Amiram Eldar, Jul 15 2022

Keywords

• nonn

Comments

Numbers k such that A355584(k) = A355584(k+1).

Equivalently, numbers k such that the largest 5-smooth divisors of k and k+1, A355582(k) and A355582(k+1), have the same sum of divisors (A000203).

For all the terms k, both k and k+1 are not squarefree: each of the two largest 5-smooth divisors, of k and k+1, cannot be squarefree, since the squarefree 5-smooth numbers are the divisors of 30 = 2*3*5 (A018255) whose values of sigma (A000203), {1, 3, 4, 6, 12, 18, 24, 72}, are not shared with sigma of any other 5-smooth number.

Apparently, all the terms are of only two types: numbers k such that A355582(k) = 16 and A355582(k+1) = 25, or numbers k such that A355582(k) = 25 and A355582(k+1) = 16. Both types are infinite sequences: The first type is the sequence of numbers of the form 2224 + 2400*m, where m is not congruent to 1 (mod 5), and the second type is the sequence of numbers of the form 175 + 2400*m, where m is not congruent to 3 (mod 5). If there are no other terms, then this sequence is a linear recurrence with a signature (1,0,0,0,0,0,0,1,-1). The question of the existence of other types is equivalent to the question of the existence of two coprime 5-smooth numbers other than 16 and 25 whose sums of divisors are equal.

Are there runs of 3 consecutive numbers with the same sum of 5-smooth divisors? There are no such runs below 5*10^10.

Examples

			175 is a term since A355584(175) = A355584(176) = 31.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000203, A002961, A018255, A355584.

Subsequence of A013929 and A068781.

Similar sequences: A002961, A064115, A064125, A293183, A306985, A333949.

Programs

Mathematica

f[p_, e_] := If[p > 5, 1, (p^(e + 1) - 1)/(p - 1)]; s[1] = 1; s[n_] := Times @@ (f @@@ FactorInteger[n]); Select[Range[10^5], s[#] == s[# + 1] &]

PARI

s(n) = (2^(valuation(n, 2) + 1) - 1) * (3^(valuation(n, 3) + 1) - 1) * (5^(valuation(n, 5) + 1) - 1) / 8;
s1 = s(1); for(k = 2, 6e4, s2 = s(k); if(s1 == s2, print1(k-1,", ")); s1 = s2);